# solution of equations by divided difference interpolaton

Divided diference interpolation can be used to obtain approximate
solutions to equations and to invert functions^{} numerically. The
idea is that, given an equation $f(y)=x$ which we want to solve
for $y$, we first take several numbers ${y}_{1},\mathrm{\dots},{y}_{n}$ and
compute ${x}_{1}\mathrm{\dots}{x}_{n}$ as ${x}_{i}=f({y}_{i})$. Then we compute the
divided differences^{} of the ${y}_{i}$’s regarded as functions of the
${x}_{i}$’s and form the divided difference series. Substituting
$x$ in this series provides an approximation to $y$.

To illustrate how this works, we will examine the transcendental equation $x+{e}^{-x}=2$. We note that $2+{e}^{-2}=2.13533$ and $1.5+{e}^{-1.5}=1.72313$, so there will be a solution between $1.5$ and $2$, likely closer to $2$ than $1.5$. Therefore, as our values of the ${y}_{i}$’s, we shall take $1.5$, $1.6$, $1.7$, $1.8$, $1.9$, $2.0$, $2.1$. We now tabulate ${x}_{i}={y}_{i}+{e}^{-{y}_{i}}$ for those values:

${y}_{i}$ | ${x}_{i}$ |

$1.5$ | $1.72313$ |

$1.6$ | $1.80190$ |

$1.7$ | $1.88268$ |

$1.8$ | $1.96530$ |

$1.9$ | $2.04957$ |

$2.0$ | $2.13533$ |

$2.1$ | $2.22246$ |

Next, we form a divided difference table of the ${y}_{i}$’s as a function of the ${x}_{i}$’s:

$$\begin{array}{cccccccc}\hfill 1.72313\hfill & \hfill 1.50000\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1.26952\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill 1.80190\hfill & \hfill 1.60000\hfill & \hfill \hfill & \hfill -0.19799\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1.23793\hfill & \hfill \hfill & \hfill 0.12082\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill 1.88268\hfill & \hfill 1.70000\hfill & \hfill \hfill & \hfill -0.16873\hfill & \hfill \hfill & \hfill -0.039609\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1.21036\hfill & \hfill \hfill & \hfill 0.10789\hfill & \hfill \hfill & \hfill 0.091553\hfill & \hfill \hfill \\ \hfill 1.96530\hfill & \hfill 1.80000\hfill & \hfill \hfill & \hfill -0.14201\hfill & \hfill \hfill & \hfill -0.077347\hfill & \hfill \hfill & \hfill -0.13457\hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1.18666\hfill & \hfill \hfill & \hfill 0.08210\hfill & \hfill \hfill & \hfill 0.024360\hfill & \hfill \hfill \\ \hfill 2.04957\hfill & \hfill 1.90000\hfill & \hfill \hfill & \hfill -0.12127\hfill & \hfill \hfill & \hfill -0.067102\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1.16604\hfill & \hfill \hfill & \hfill 0.05930\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill 2.13533\hfill & \hfill 2.00000\hfill & \hfill \hfill & \hfill -0.10602\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1.14771\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill 2.22246\hfill & \hfill 2.10000\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \end{array}$$ |

From this table, we form the series

$1.50000$ | $+1.26952(x-1.72313)-0.19799(x-1.72313)(x-1.80190)$ | ||

$+0.12082(x-1.72313)(x-1.80190)(x-1.88268)$ | |||

$-0.039609(x-1.72313)(x-1.80190)(x-1.88268)(x-1.96530)$ | |||

$+0.091553(x-1.72313)(x-1.80190)(x-1.88268)(x-1.96530)(x-2.04957)$ | |||

$-0.13457(x-1.72313)(x-1.80190)(x-1.88268)(x-1.96530)(x-2.04957)(x-2.13533)$ |

Substituting $2.00000$ for $x$, we obtain $1.84140$. Given that

$$ |

this answer is correct to all $5$ decimal places.

In the presentation above, we tacitly assumed that there was a
solution to our equation and focussed our attention on finding
that answer numerically. To complete^{} the treatment we will
now show that there indeed exists a unique solution to the
equation $x+{e}^{-x}=2$ in the interval $(0,\mathrm{\infty})$.

Existence follows from the intermediate value theorem. As noted above,

$$ |

Since $x+{e}^{-x}$ depends continuously on $x$, it follows that there exists $x\in (1.5,2)$ such that $x+{e}^{-x}=2$.

As for uniqueness, note that the derivative^{} of $x+{e}^{-x}$
is $1-{e}^{-x}$. When $x>0$, we have $$, or
$1-{e}^{-x}>0$. Hence, $x+{e}^{-x}$ is a strictly increaing
function of $x$, so there can be at most one $x$ such that
$x+{e}^{-x}=2$.

Title | solution of equations by divided difference interpolaton |
---|---|

Canonical name | SolutionOfEquationsByDividedDifferenceInterpolaton |

Date of creation | 2013-03-22 16:49:22 |

Last modified on | 2013-03-22 16:49:22 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 16 |

Author | rspuzio (6075) |

Entry type | Application |

Classification | msc 39A70 |